Answer:
Discontinuity at (−4, −3), zero at (−1, 0)
Step-by-step explanation:
The expression simplifies to ...
[tex]\dfrac{x^2+5x+4}{x+4}=\dfrac{(x+4)(x+1)}{(x+4)}=x+1 \qquad\text{$x\ne -4$}[/tex]
This is undefined at x=-4, hence there is a discontinuity there, and is zero for x=-1.
There is a discontinuity at x=-4, and a zero at (-1, 0).
